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3 years ago

A 0.450-kg soccer ball has a kinetic energy of 119 J. What is the velocity of the soccer ball?

Physics

1 answer:

Elis [28]3 years ago

3 0

Answer: 23m/s

Explanation:

Given that:

Mass of soccer ball = 0.450kg

kinetic energy = 119 J

velocity of the soccer ball = ?

Kinetic energy is the energy possessed by a moving object. It is measured in joules, and depends on the mass (m) of the object and the velocity (v) by which it moves

i.e K.E = 1/2mv^2

119J = 1/2 x 0.450kg x v^2

119J = 0.225kg x v^2

v^2 = 119J / 0.225kg

v^2 = 528.9

To get the value of velocity, find the square root of v

v = √528.9

v = 22.997m/s (Round result to 23)

Thus, the velocity of the soccer ball is 23m/s

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Which property helps to explain differences in the specific heat capacities of

tatyana61 [14]

Answer:

D. Forces between molecules

Explanation:

Specific heat capacity of water can be defined as the amount of heat a gram of water must lose or absorb in order to change its temperature by a degree Celsius. It is measured in Joules per kilogram per degree Celsius (J/kg°C). Generally, the specific heat capacity of water is 4.182J/kg°C and is the highest among liquids.

Mathematically, the specific heat capacity of a substance is given by the formula;

$c = \frac {Q}{mdt}$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

dt represents the change in temperature.

Cohesion is a property of water and it typically refers to the attraction between molecules of water which holds them together.

In Science, the property which helps to explain differences in the specific heat capacities of two substances is the forces between molecules.

This ultimately implies that, the more closely bonded the atoms of a substance are, the higher or greater would be the substance's specific heat capacity. Thus, it varies for the various states of matter i.e solid, liquid and gas.

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3 years ago

What is the magnitude of the angular momentum relative to the origin of the 100 g particle in the figure(figure 1 ? express your

torisob [31]

<span>Radius distance from origin to particle = √ (2²+1²) = √5 m = R
I = MR² = (0.200)(5) = 1.00 kg-m²
Θ = arctan 2/1 = 63.4° = R's angle CCW from horizontal
V = 3.0 m/s
V component that is at 90° to R = 3.0(sin 90°- 63.4°) = 3.0(sin 26.6°) = 1.3433 m/s
w = [V component / R] = 1.3433/√5 = 0.601 rad/s
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3 years ago

One of the harmonic frequencies of tube A with two open ends is 576 Hz. The next-highest harmonic frequency is 648 Hz. (a) What

balu736 [363]

(a) 288 Hz

The difference between any two harmonics of an open-end tube is equal to the fundamental frequency, $f_1$ (first harmonic):

$f_{n+1}-f_n = f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 576 Hz\\f_{n+1}=648 Hz$

So the fundamental frequency is:

$f_1 = 648 Hz-576 Hz=72 Hz$

Now we know that one of the the harmonics is $f_n=216 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+f_1 = 216 Hz+72 Hz=288 Hz$

(b) n=4

The frequency of the nth-harmonic is an integer multiple of the fundamental frequency:

$f_n=n f_1$ (2)

Since we know $f_n = 288 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{f_n}{f_1}=\frac{288 Hz}{72 Hz}=4$

(c) 4445 Hz

For a closed pipe (only one end is open), the situation is a bit different, because only odd harmonics are allowed. This means that the frequency of the nth-harmonic is an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)